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Ch. 4 – 1 st Law of Thermodynamics

Ch. 4 – 1 st Law of Thermodynamics. 1 st Law and Enthalpy As noted earlier, Q = dU + W is the simplest expression of the 1 st Law. When applied to an isolated system ( dU = 0), it expresses the law of conservation of energy

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Ch. 4 – 1 st Law of Thermodynamics

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  1. Ch. 4 – 1st Law of Thermodynamics • 1st Law and Enthalpy • As noted earlier, Q = dU + Wis the simplest expression of the 1st Law. When applied to an isolated system (dU = 0), it expresses the law of conservation of energy • Introduce another State Function (besides U), called enthalpy, H (=exact differential) • This leads to two forms of the 1st Law

  2. Ch. 4 – 1st Law of Thermodynamics • 1st Law and Enthalpy • These have integral equivalents and can be referred to unit mass • Note, the above expressions assume the process to be quasi static (sequence of states that are infinitesimally close to equilibrium), if p stands for the internal pressure of the system. Otherwise, p is the external pressure exerted on the system.

  3. Ch. 4 – 1st Law of Thermodynamics • Heat Capacities • Consider a homogeneous system of constant composition • Write dU (dH) as a total differential of the independent variables X and Y (for our purposes, X and Y could be any pair of p, V, and T). Then • From the 1st Law, Q = dU + pdV, we can substitute to get where X and Y are Tand V, respectively.

  4. Ch. 4 – 1st Law of Thermodynamics • Heat Capacities • From Q = dH – Vdpwe can substitute for dH to get where X and Y are T and p, respectively. • For a process of heating at V = const we get the thermal capacity at constant V or per unit mass we can write

  5. Ch. 4 – 1st Law of Thermodynamics • Heat Capacities • For a process of heating at p = const we get the thermal capacity at constant p or per unit mass we can write • Can also calculate heat of change of V and p at T = const • Not very useful, but given for completeness

  6. Ch. 4 – 1st Law of Thermodynamics • Calculation of Internal Energy and Enthalpy • Our equations for thermal capacities CV and Cp can be integrated directly for processes with V = const and p = const, respectively to find U and H, if CV and Cp are known as functions of T • CV and Cp, from experiment, are usually polynomials in T

  7. Ch. 4 – 1st Law of Thermodynamics • Calculation of Internal Energy and Enthalpy • The experimental design of Joule’s proof that U=U(T) (pg 34, Tsonis) • Consider 2 rigid vessels linked by a connection with a stopcock. One contains gas, the other evacuated. (both placed in a calorimeter with thermometer) • Stopcock opened, gas in 1 expands to occupy both vessels, until pressure everywhere is equalized. • Temperature measurements show ∆T=0  that system exchanges no heat with environment Q = 0  ∆U=-W; since the volume of the container (the system) does not change  no work done,, W = 0, and ΔU = 0. • How about the gas in compartment A? Its volume increases from A  A+B. Since ∆T=0 and ΔU = 0  variation in volume at T=const produces no variation in energy  Since p changed during process, we have U = U(T) only and partial derivative used above are total derivatives.

  8. Ch. 4 – 1st Law of Thermodynamics • Calculation of Internal Energy and Enthalpy • Notes on this experiment • When experiment done carefully, small heat exchange was found (Joule-Thomson effect), which vanishes for ideal gas behavior • As gas confined in vessel 1 expands into 2, work is done by some portions of gas against others while volumes change (as molecules enter 2, they are effected by molecules following) • These are internal transfers that are not included in W • This shows the importance of defining system carefully and clearly when considering a thermodynamic process • In this case, the system is all the gas contained in both vessels (initially one is empty), whose total volume (V1 + V2) does not change

  9. Ch. 4 – 1st Law of Thermodynamics • More on Heat Capacities • As noted above, since U = U(T) we have CV = dU/dTand cv = du/dT • We can write H = U + pV = U + nR*T = H(T) leading to Cp = dH/dTand cp = dh/dT • Since we are only interested in differences in internal energy and enthalpy, we can set the integration constant to 0 giving: U  CVT, H  CpT, u  cvT, and h  cpT

  10. Ch. 4 – 1st Law of Thermodynamics • More on Heat Capacities • Since we have CV = dU/dT, Cp = dH/dTand H = U + pV = U + nR*T, we have leading to Cp – CV = nR* cp – cv = Rrecalling that n = m/M. • As mentioned earlier, heat capacities for all gases can be measured and the coefficients for the polynomial expansion can be determined (C =  + T + T2 + …)

  11. Ch. 4 – 1st Law of Thermodynamics • More on Heat Capacities • For simple gases like N2, O2, and Ar, the experimental data are nearly constant for all temperatures and pressures of interest, so the temperature variation is not considered. • From earlier we have, for monatomic gases, the total internal energy is U = (3/2)NkT, which leads to CV = (3/2)nR*and cv = (3/2)R. Similarly, Cp = (5/2)nR*and cp = (5/2)R. • For diatomic gases, where there are more degrees of freedom, so we get CV = (5/2)nR*and cv = (5/2)R. Similarly, Cp = (7/2)nR* and cp = (7/2)R. • The ratios Cp/CV = cp/cv = .

  12. Ch. 4 – 1st Law of Thermodynamics • More on Heat Capacities • Dry air is considered to be a diatomic gas, so the second form applies. • The ratio, cp/cv =  = 1.4. • We then attach the subscript, d, to the specific heats to indicate dry air. This leads to: cvd = 718 J kg-1 K-1 cpd = 1005 J kg-1 K-1 Rd = cpd – cvd = 287(.05) J kg-1 K-1.

  13. Ch. 4 – 1st Law of Thermodynamics • More on Heat Capacities • The table below shows the values of cpd for various temperatures and pressures. Note the slight variation.

  14. Ch. 4 – 1st Law of Thermodynamics • More Forms of the 1st Law • Using the above expressions for (specific) heat capacities, we get more useful forms of the 1st Law, two of which are particularly useful and

  15. Ch. 4 – 1st Law of Thermodynamics • Special Cases • For an isothermal transformation • For an isochoric (constant volume) transformation • For an isobaric transformation

  16. Example 4.4:Consider an insolated box with two compartments A and B. Both • Compartments contain a mono-atomic ideal gas and they are separated by a wall • that on one hand does not allow any heat through it but on the other hand is • flexible enough to ensure equalization of pressure in both compartments. • The initial conditions for both compartments are the same: Ti=273 K, Vi=1000 cm3, • pi=1 atm = 101300 Pa. Then by means of an electrical resistance, heat is supplied to • the gas in A until its pressure becomes ten times its initial pressure. Estimate: • The final temperature of the gas in B; • The work performed on the gas in B; • The final temperature of the gas in A; • The heat absorbed by the gas in A. • From Poisson Eqn: • For mono-atomic gas:

  17. Example 4.4: continue (b) The work performed on the gas in B is given by:

  18. Example 4.4: continue (c) Final volume in B is 250 cm3 volume in A is 1750 cm3. (d) The heat absorbed by the gas in A is given by:

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